Why Do Stars Flicker but Not Planets?

  • Thread starter Thread starter mohdfasieh
  • Start date Start date
  • Tags Tags
    Genius
AI Thread Summary
Stars flicker due to atmospheric effects that cause light to bend and change direction, while planets appear more stable because they have a larger angular size and emit light as a disk rather than a point. The flickering is perceived as changes in intensity, but it primarily results from variations in the atmosphere's density. When observing planets, their larger size ensures that some light consistently reaches the observer's eye, unlike the smaller, point-like appearance of stars. The discussion also touches on the calculation of photon entry into the eye, emphasizing the importance of pupil size and observation time. Ultimately, the flicker effect is unrelated to the number of photons received, which remains significant for both stars and planets.
mohdfasieh
Messages
26
Reaction score
0
HELLO,

CAn any genius solve this question.I have tried my best but iam unable to get the answer.


QUESTION 1 ; "Stars are know to flicker.Estimate the number of photons entering the eye of an observer when he looks at the star of first apparent visual magnitude.Such a star produce flux on the surface of the Earth of 10 (exp) -6 lumens/m^2 .One lumen is 0.0016 watts.Star Aldebaran is an example . Why do stars flicker,Not the palnets or very little.




QUESTION 2 :In positron atom both the electron and the positron mive with the center of amss that is sationary .Calculate the ground state energy of the positron atom.
 
Physics news on Phys.org
I think stars flicker more than planets as they are lot further away.The intermediary medium has varied properties so they appear as blinking.
 
But the "intermediary medium" is, primarily, the atmospher which is just a deep for planets as for stars. The actual reason is that planets have a disk while stars do not. If you look at planets in a telescope, all those that are visible with the naked eye, you will see a disk. If you look at stars in a telescope, no matter how how powerful the telescope, you see a single point. Even with the naked eye, you see a tiny disk, as opposed to a point of light. A disk is not as easily covered or "warped" by a change in the density of the atmosphere as a single point.

Oh, and to find how many photons enter the eye, you will need to know the size of the pupil as well as the time period involved.
 
Stars flicker because of atmospheric effects only. Think about what the sun looks like when seen from the bottom of a pool of water with ripples in the surface. The flicker of stars is similar, but instead of the varying surface of a pool, the reason light gets bent in various directions in the atmosphere is due to the fact that air does not have uniform density.

Sometimes, if the air is very still, the stars will not flicker.

The flicker can be described two different ways. What your eye perceives is a change in intensity. But what is really going on is a change in direction. Sometimes the intensity is lower because the light changed direction and missed you.

Both a star and a planet are "disk" sources of light, but a star is far smaller. A planet is so much larger (in angular area) that there is always some light that makes it to your eye. Another way of putting this is that the amount of change to the direction of light due to flicker is LESS than the angular size of the planet. Thus the planet stays in view and appears stable.

The flicker effect has nothing to do with the number of photons entering the eye. In both the case of a planet or a first magnitude star, you will find that the number of photons received by the eye is a fairly large number, per second.

Carl
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
14
Views
6K
Replies
16
Views
24K
Back
Top